Let $h$ be a differentiable function with $h(-6)=2$ and $h'(-6)=-1$. What is the value of the approximation of $h(-6.2)$ using the function's local linear approximation at $x=-6$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1.9$ (Choice B) B $2$ (Choice C) C $2.1$ (Choice D) D $2.2$
Answer: The local linear approximation of $h$ at $x=-6$ is achieved using the equation of the line tangent to $h$ at $x=-6$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=h'(-6)(x+6)+h(-6)$. Plugging $h(-6)=2$ and $h'(-6)=-1$, we obtain $L(x)=-1(x+6)+2$. To approximate $h(-6.2)$, all we need is to plug $x=-6.2$ into $L(x)$. $\begin{aligned} L(-6.2)&=-1(-6.2+6)+2 \\\\ &=-1(-0.2)+2 \\\\ &=2.2 \end{aligned}$ In conclusion, the approximation of $h(-6.2)$ using the function's local linear approximation at $x=-6$ is $2.2$.